4W
To solve this problem, we need to analyze the energy dissipation rate before and after the wire is cut.
Initial Energy Dissipation Rate: The power P dissipated by a resistor is given by:
\( P = \frac{V^2}{R} \),
where: - V is the voltage across the resistor, and R is the resistance.
Initially, with resistance R, the energy dissipation rate is:
\( W = P = \frac{V^2}{R} \).
After Cutting the Wire: When the wire is cut into two halves, each half has a resistance of:
\( R' = \frac{R}{2} \).
Connecting in Parallel: When these two halves are connected in parallel, the equivalent resistance Req is given by:
\( \frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R/2} = \frac{4}{R} \).
Thus, the equivalent resistance is:
\( R_{eq} = \frac{R}{4} \).
New Energy Dissipation Rate: The new power P' dissipated in the circuit with the new resistance Req is:
\( P' = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = \frac{4V^2}{R} \).
Comparing Power Dissipation Rates: Since \( W = \frac{V^2}{R} \) from the original circuit, we can relate the new power:
\( P' = 4W \).
The following are the graphs of potential barrier versus width of the depletion region for a p-n junction diode.
Which of the following is correct?
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
A P-N junction is an interface or a boundary between two semiconductor material types, namely the p-type and the n-type, inside a semiconductor.
in p-n junction diode two operating regions are there:
There are three biasing conditions for p-n junction diode are as follows: